A) 0.55 atm
B) 0.11 atm
C) 1 atm
D) 0.12 atm
Correct Answer: C
Solution :
No. of moles of lighter gas \[=\frac{m}{4}\] No. of moles of heavier gas \[=\frac{m}{40}\] Total no. of moles \[=\frac{m}{4}+\frac{m}{40}=\frac{11m}{40}\] Mole fraction of lighter gas \[=\frac{\frac{m}{4}}{\frac{11m}{40}}=\frac{10}{11}\] Partial pressure due to lighter gas \[={{P}_{o}}\times \frac{10}{11}\] \[=1.1\times \frac{10}{11}=1atm.\]
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